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-Cái này áp dụng hằng đẳng thức số 3 á.
\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x-5+x+2\right)\left(2x-5-x-2\right)=0\)
\(\Leftrightarrow\left(3x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
Vậy...
Bài eassy
\(\left(2x-5\right)^2-\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
Vậy.....................
a) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Leftrightarrow x-2=0\) (Vì: \(x^2+4x+6>0\) )
\(\Leftrightarrow x=2\)
b) \(2x^3+x^2-6x=0\)
\(\Leftrightarrow x\left(2x^2+x-6\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2+4x\right)-\left(3x+6\right)\right]=0\)
\(\Leftrightarrow x\left[2x\left(x+2\right)-3\left(x+2\right)\right]=0\)
\(\Leftrightarrow x\left(x+2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+2=0\\2x-3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\\x=\frac{3}{2}\end{array}\right.\)
c) \(4x^2+4xy+x^2-2x+1+y^2=0\)
\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2=0\)
\(\Leftrightarrow\begin{cases}2x+y=0\\x-1=0\end{cases}\)\(\Leftrightarrow\begin{cases}y=-2\\x=1\end{cases}\)
4, \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(=5x^2+5\ge5\)
Dấu "=" xảy ra khi x=0
5,\(A=4-x^2+2x=5-\left(x^2-2x+1\right)=5-\left(x-1\right)^2\le5\)
Dấu "=" xảy ra khi x=1
\(B=4x-x^2=4-\left(x^2-4x+4\right)=4-\left(x-2\right)^2\le4\)
Dấu "=" xảy ra khi x=2
a) \(x^3+2x^2-4x+1\)
\(=\left(x^3+3x^2-x\right)-\left(x^2+3x-1\right)\)
\(=x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x-1\right)\left(x^2+3x-1\right)\)
c) cho da thuc P(x) =2x^4-7x^3 -2x^2 +13x +6? | Yahoo Hỏi & Đáp
Tham khảo
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-4\left(x^2-2x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)-\left(2x^2-4x+2\right)^2=0\)
\(\Leftrightarrow\left(x+1-2x^2+4x-2\right)\left(x+1+2x^2-4x+2\right)=0\)
\(\Leftrightarrow\left(-2x^2+5x-1\right)\left(2x^2-3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2+5x-1=0\\2x^2-3x+3=0\left(loai\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{17}}{4}\\x=\dfrac{5-\sqrt{17}}{4}\end{matrix}\right.\)